2011 USAJMO Problems/Problem 3

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Problem

For a point $P = (a, a^2)$ in the coordinate plane, let $\ell(P)$ denote the line passing through $P$ with slope $2a$. Consider the set of triangles with vertices of the form $P_1 = (a_1,  a_1^2)$, $P_2 = (a_2, a_2^2)$, $P_3 = (a_3, a_3^2)$, such that the intersections of the lines $\ell(P_1)$, $\ell(P_2)$, $\ell(P_3)$ form an equilateral triangle $\triangle$. Find the locus of the center of $\triangle$ as $P_1P_2P_3$ ranges over all such triangles.

Solutions

Solution 1

Note that the lines $l(P_1), l(P_2), l(P_3)$ are \[y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,\] respectively. It is easy to deduce that the three points of intersection are \[\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).\] The slopes of each side of this equilateral triangle are \[2a_1,2a_2,2a_3,\] and we want to find the locus of \[\left(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_3a_1}{3}\right).\] Define the three complex numbers $w_n = 1+2a_ni$ for $n=1,2,3$. Then note that the slope - that is, the imaginary part divided by the real part - of all $w_n^3$ is constant, say it is $k$. Then for $n=1,2,3$,

\begin{align*} \frac{\Im(w_n^3)}{\Re(w_n^3)} &= \frac{\Im((1+2a_ni)^3)}{\Re((1+2a_ni)^3)}\\ &= \frac{\Im(1+6a_ni-12a_n^2-8a_n^3i)}{\Re(1+6a_ni-12a_n^2-8a_n^3i)}\\ &= \frac{6a_n-8a_n^3}{1-12a_n^2}\\ &= k.\\ \end{align*}

Rearranging, we get that\[8a_n^3 -12ka_n^2-6a_n+k=0,\]or\[a_n^3-\frac{3ka_n^2}2-\frac{3a_n}4+\frac k8=0.\]Note that this is a cubic, and the roots are $a_1,a_2$ and $a_3$ which are all distinct, and so there are no other roots. Using Vieta's, we get that \[a_1+a_2+a_3=\frac{3k}2,\]and\[a_1a_2+a_2a_3+a_3a_1=-\frac34.\] Obviously all values of $k$ are possible, and so our answer is the line \[\boxed{y=-\frac{1}{4}.}\] $\blacksquare$

~ cocohearts

Solution 2

Note that all the points $P=(a,a^2)$ belong to the parabola $y=x^2$ which we will denote $p$. This parabola has a focus $F=\left(0,\frac{1}{4}\right)$ and directrix $y=-\frac{1}{4}$ which we will denote $d$. We will prove that the desired locus is $d$.

First note that for any point $P$ on $p$, the line $\ell(P)$ is the tangent line to $p$ at $P$. This is because $\ell(P)$ contains $P$ and because $[\frac{d}{dx}] x^2=2x$. If you don't like calculus, you can also verify that $\ell(P)$ has equation $y=2a(x-a)+a^2$ and does not intersect $y=x^2$ at any point besides $P$. Now for any point $P$ on $p$ let $P'$ be the foot of the perpendicular from $P$ onto $d$. Then by the definition of parabolas, $PP'=PF$. Let $q$ be the perpendicular bisector of $\overline{P'F}$. Since $PP'=PF$, $q$ passes through $P$. Suppose $K$ is any other point on $q$ and let $K'$ be the foot of the perpendicular from $K$ to $d$. Then in right $\Delta KK'P'$, $KK'$ is a leg and so $KK'<KP'=KF$. Therefore $K$ cannot be on $p$. This implies that $q$ is exactly the tangent line to $p$ at $P$, that is $q=\ell(P)$. So we have proved Lemma 1: If $P$ is a point on $p$ then $\ell(P)$ is the perpendicular bisector of $\overline{P'F}$.

We need another lemma before we proceed. Lemma 2: If $F$ is on the circumcircle of $\Delta XYZ$ with orthocenter $H$, then the reflections of $F$ across $\overleftrightarrow{XY}$, $\overleftrightarrow{XZ}$, and $\overleftrightarrow{YZ}$ are collinear with $H$.

Proof of Lemma 2: Say the reflections of $F$ and $H$ across $\overleftrightarrow{YZ}$ are $C'$ and $J$, and the reflections of $F$ and $H$ across $\overleftrightarrow{XY}$ are $A'$ and $I$. Then we angle chase $\angle JYZ=\angle HYZ=\angle HXZ=\angle JXZ=m(JZ)/2$ where $m(JZ)$ is the measure of minor arc $JZ$ on the circumcircle of $\Delta XYZ$. This implies that $J$ is on the circumcircle of $\Delta XYZ$, and similarly $I$ is on the circumcircle of $\Delta XYZ$. Therefore $\angle C'HJ=\angle FJH=m(XF)/2$, and $\angle A'HX=\angle FIX=m(FX)/2$. So $\angle C'HJ = \angle A'HX$. Since $J$, $H$, and $X$ are collinear it follows that $C'$, $H$ and $A'$ are collinear. Similarly, the reflection of $F$ over $\overleftrightarrow{XZ}$ also lies on this line, and so the claim is proved.

Now suppose $A$, $B$, and $C$ are three points of $p$ and let $\ell(A)\cap\ell(B)=X$, $\ell(A)\cap\ell(C)=Y$, and $\ell(B)\cap\ell(C)=Z$. Also let $A''$, $B''$, and $C''$ be the midpoints of $\overline{A'F}$, $\overline{B'F}$, and $\overline{C'F}$ respectively. Then since $\overleftrightarrow{A''B''}\parallel \overline{A'B'}=d$ and $\overleftrightarrow{B''C''}\parallel \overline{B'C'}=d$, it follows that $A''$, $B''$, and $C''$ are collinear. By Lemma 1, we know that $A''$, $B''$, and $C''$ are the feet of the altitudes from $F$ to $\overline{XY}$, $\overline{XZ}$, and $\overline{YZ}$. Therefore by the Simson Line Theorem, $F$ is on the circumcircle of $\Delta XYZ$. If $H$ is the orthocenter of $\Delta XYZ$, then by Lemma 2, it follows that $H$ is on $\overleftrightarrow{A'C'}=d$. It follows that the locus described in the problem is a subset of $d$.

Since we claim that the locus described in the problem is $d$, we still need to show that for any choice of $H$ on $d$ there exists an equilateral triangle with center $H$ such that the lines containing the sides of the triangle are tangent to $p$. So suppose $H$ is any point on $d$ and let the circle centered at $H$ through $F$ be $O$. Then suppose $A$ is one of the intersections of $d$ with $O$. Let $\angle HFA=3\theta$, and construct the ray through $F$ on the same halfplane of $\overleftrightarrow{HF}$ as $A$ that makes an angle of $2\theta$ with $\overleftrightarrow{HF}$. Say this ray intersects $O$ in a point $B$ besides $F$, and let $q$ be the perpendicular bisector of $\overline{HB}$. Since $\angle HFB=2\theta$ and $\angle HFA=3\theta$, we have $\angle BFA=\theta$. By the inscribed angles theorem, it follows that $\angle AHB=2\theta$. Also since $HF$ and $HB$ are both radii, $\Delta HFB$ is isosceles and $\angle HBF=\angle HFB=2\theta$. Let $P_1'$ be the reflection of $F$ across $q$. Then $2\theta=\angle FBH=\angle C'HB$, and so $\angle C'HB=\angle AHB$. It follows that $P_1'$ is on $\overleftrightarrow{AH}=d$, which means $q$ is the perpendicular bisector of $\overline{FP_1'}$.

Let $q$ intersect $O$ in points $Y$ and $Z$ and let $X$ be the point diametrically opposite to $B$ on $O$. Also let $\overline{HB}$ intersect $q$ at $M$. Then $HM=HB/2=HZ/2$. Therefore $\Delta HMZ$ is a $30-60-90$ right triangle and so $\angle ZHB=60^{\circ}$. So $\angle ZHY=120^{\circ}$ and by the inscribed angles theorem, $\angle ZXY=60^{\circ}$. Since $ZX=ZY$ it follows that $\Delta ZXY$ is and equilateral triangle with center $H$.

By Lemma 2, it follows that the reflections of $F$ across $\overleftrightarrow{XY}$ and $\overleftrightarrow{XZ}$, call them $P_2'$ and $P_3'$, lie on $d$. Let the intersection of $\overleftrightarrow{YZ}$ and the perpendicular to $d$ through $P_1'$ be $P_1$, the intersection of $\overleftrightarrow{XY}$ and the perpendicular to $d$ through $P_2'$ be $P_2$, and the intersection of $\overleftrightarrow{XZ}$ and the perpendicular to $d$ through $P_3'$ be $P_3$. Then by the definitions of $P_1'$, $P_2'$, and $P_3'$ it follows that $FP_i=P_iP_i'$ for $i=1,2,3$ and so $P_1$, $P_2$, and $P_3$ are on $p$. By lemma 1, $\ell(P_1)=\overleftrightarrow{YZ}$, $\ell(P_2)=\overleftrightarrow{XY}$, and $\ell(P_3)=\overleftrightarrow{XZ}$. Therefore the intersections of $\ell(P_1)$, $\ell(P_2)$, and $\ell(P_3)$ form an equilateral triangle with center $H$, which finishes the proof. --Killbilledtoucan

Solution 3

Note that the lines $l(P_1), l(P_2), l(P_3)$ are \[y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,\] respectively. It is easy to deduce that the three points of intersection are \[\left(\frac{a_1+a_2}{2},a_1a_2\right),\left(\frac{a_2+a_3}{2},a_2a_3\right), \left(\frac{a_3+a_1}{2},a_3a_1\right).\] The slopes of each side of this equilateral triangle are \[2a_1,2a_2,2a_3,\] and we want to find the locus of \[\left(\frac{a_1+a_2+a_3}{3},\frac{a_1a_2+a_2a_3+a_3a_1}{3}\right).\] We know that \[2a_1=\tan(\theta), 2a_2=\tan (\theta + 120), 2a_3=\tan (\theta-120)\] for some $\theta.$ Therefore, we can use the tangent addition formula to deduce \[\frac{a_1+a_2+a_3}{3}=\frac{\tan(\theta)+\tan (\theta + 120)+\tan (\theta-120)}{6}=\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}\] and \begin{align*} \frac{a_1a_2+a_2a_3+a_3a_1}{3}&=\frac{\tan\theta (\tan(\theta-120)+\tan(\theta+120))+\tan(\theta-120)\tan(\theta+120)}{12}\\ &=\frac{9\tan^2\theta-3}{12(1-3\tan^2\theta)}\\ &=-\frac{1}{4}.\end{align*} Now we show that $\frac{a_1+a_2+a_3}{3}$ can be any real number. Let's say \[\frac{3\tan\theta-\tan^3\theta}{2-6\tan^2\theta}=k\] for some real number $k.$ Multiplying both sides by $2-\tan^2\theta$ and rearranging yields a cubic in $\tan\theta.$ Clearly this cubic has at least one real solution. As $\tan \theta$ can take on any real number, all values of $k$ are possible, and our answer is the line \[\boxed{y=-\frac{1}{4}.}\] Of course, as the denominator could equal 0, we must check $\tan \theta=\pm \frac{1}{\sqrt{3}}.$ \[3\tan \theta-\tan^3\theta=k(2-6\tan^2\theta).\] The left side is nonzero, while the right side is zero, so these values of $\theta$ do not contribute to any values of $k.$ So, our answer remains the same. $\blacksquare$ ~ Benq

See also

2011 USAJMO (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6
All USAJMO Problems and Solutions

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